On what days of week during the Gregorian period can a century begin?

This problem's question is: On what days of week during the Gregorian period can a century begin?

For this problem, a century begins on January 1, CC01, where centuries number from 1, which produces a CC number one less than the century number.

I suggest that you start your analysis from Monday, January 1, 2001, which is such a year and also starts a four Gregorian century cycle. The Gregorian leap year rules can be found here .

Hint: check the length of a four Gregorian century cycle modulus 7.

Verifiable fact: January 1, 2001 was, in fact a Monday.

Monday, Tuesday, Wednesday, Thursday, Saturday Monday, Wednesday, Sunday Monday, Tuesday, Thursday, Saturday Monday, Tuesday, Wednesday, Thursday, Friday Monday, Wednesday, Thursday, Friday, Saturday

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2 solutions

Kyle T
Jun 4, 2019

similar to my code for another one of your problems
we can see that the only days of the week found are Monday, Tuesday, Thursday and Saturday

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<?php
$results = array();
$days = array('sun','mon','tue','wed','thu','fri','sat');
for($i=0;$i<=6;$i++){
    $results[$days[$i]] = 0;
}
for($y=2001;$y<=2400;$y+=100){
    $time = mktime(0,0,0,1,1,$y);
    $results[$days[date('w',$time)]]++;
}
echo '<pre>'.print_r($results,true).'</pre>';
/*
Array
(
    [sun] => 0
    [mon] => 1
    [tue] => 1
    [wed] => 0
    [thu] => 1
    [fri] => 0
    [sat] => 1
)
*/
?>

January 1, 2001, was in fact a Monday.

An ordinary Gregorian century has 76 ordinary and 24 leap years. 76 × 365 + 24 × 366 m o d 7 = 5 76\times 365+24\times 366 \mod 7=5 .

A leap Gregorian century has 75 ordinary and 25 leap years. 75 × 365 + 25 × 366 m o d 7 = 6 75\times 365+25\times 366 \mod 7=6 .

Note, 3 × 5 + 6 m o d 7 = 0 3\times 5+6 \mod 7=0 , which proves the four Gregorian century cycle repeats the day of week pattern.

Five days after Monday is Saturday. Five days after Saturday is Thursday. Five days after Thursday is Tuesday .

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